Question

2. Allan, a Grade 12 senior high school student, wants to estimate the average number ofstudents who will pursue collage degree in a certain school. How many sample sizes does heneed, if he plans to use 98% confidence, 0.5 as the margin of error, and a standard deviation of5.​

Answer 1

Given:

The margin of error E = ± 0.5 %

standard deviation σ = 5

Z = 1.96 when confidence is 98%

To find:

sample sizes = N =?

Step-by-step explanation:

In a normal distribution table, the necessary sample size is calculated as follows:

$\displaystyle N = \frac{Z^2\times \sigma \times (1 - \sigma)}{E^2}$

where

N = Necessary Sample Size

Z = statistics when confidence is 98%

σ = standard deviation

E = margin of error

Put the given values in the above equation

$\displaystyle N = \frac{(1.96)^2 \times[ 0.5 - (1 - 0.5]}{( 0.05)^2}$

N = 384

∴ 384 sample sizes are needed.